To find the sum `sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7)`, we follow the following method.
Put `7theta = 2npi`, where `n ` is any integer. Then
`" " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta`
This means that `sin theta` takes the values `0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7)`.
From Eq. (i), we now get
`" " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta `
or `4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3)`
Rejecting the value `sin theta =0`, we get
`" " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3`
or ` 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2)`
or `16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta)`
`" " = 16 sin ^(4) theta - 24 sin ^(2) theta +9`
or `" " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0`
This is cubic in `sin^(2) theta` with the roots `sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7)`.
The sum of these roots is
`" " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4)`.
The value of `(tan^(2)""(pi)/(7) + tan^(2)""(2pi)/(7) + tan^(2)""(3pi)/(7))/(cot^(2)""(pi)/(7) + cot^(2)""(2pi)/(7) + cot^(2)""(3pi)/(7))` is

Options

A. 7
B. `35//3`
C. `21//5`
D. none of these

Explanation

To solve the problem, we need to find the value of

\[ \frac{\tan^2\left(\frac{\pi}{7}\right) + \tan^2\left(\frac{2\pi}{7}\right) + \tan^2\left(\frac{3\pi}{7}\right)}{\cot^2\left(\frac{\pi}{7}\right) + \cot^2\left(\frac{2\pi}{7}\right) + \cot^2\left(\frac{3\pi}{7}\right)}. \]

### Step 1: Define the angles
Let \( \theta = \frac{\pi}{7} \). Therefore, we need to evaluate:

\[ \frac{\tan^2(\theta) + \tan^2(2\theta) + \tan^2(3\theta)}{\cot^2(\theta) + \cot^2(2\theta) + \cot^2(3\theta)}. \]

### Step 2: Rewrite cotangent in terms of tangent
Recall that \( \cot^2(x) = \frac{1}{\tan^2(x)} \). Thus, we can rewrite the denominator:

\[ \cot^2(\theta) + \cot^2(2\theta) + \cot^2(3\theta) = \frac{1}{\tan^2(\theta)} + \frac{1}{\tan^2(2\theta)} + \frac{1}{\tan^2(3\theta)}. \]

### Step 3: Let \( z = \tan^2(\theta) \)
Let \( z_1 = \tan^2(\theta) \), \( z_2 = \tan^2(2\theta) \), and \( z_3 = \tan^2(3\theta) \). The expression now becomes:

\[ \frac{z_1 + z_2 + z_3}{\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3}}. \]

### Step 4: Simplify the denominator
The denominator can be rewritten as:

\[ \frac{z_1 + z_2 + z_3}{\frac{z_2 z_3 + z_1 z_3 + z_1 z_2}{z_1 z_2 z_3}} = \frac{(z_1 + z_2 + z_3) z_1 z_2 z_3}{z_2 z_3 + z_1 z_3 + z_1 z_2}. \]

### Step 5: Use the sum of roots
From the previous analysis, we know that the sum of the roots \( z_1 + z_2 + z_3 \) is given by the equation derived from the cubic equation in \( z^2 \):

\[ z^6 - 21z^4 + 35z^2 - 7 = 0. \]

From Vieta's formulas, the sum of the roots \( z_1 + z_2 + z_3 = 21 \).

### Step 6: Find the product of the roots
The product of the roots \( z_1 z_2 z_3 \) is given by \( \frac{-(-7)}{1} = 7 \).

### Step 7: Find the sum of the products of the roots taken two at a time
The sum of the products of the roots taken two at a time \( z_1 z_2 + z_2 z_3 + z_1 z_3 = 35 \).

### Step 8: Substitute into the expression
Now substituting these values into our expression, we have:

\[ \frac{21 \cdot 7}{35} = \frac{147}{35} = \frac{21}{5}. \]

### Final Answer
Thus, the value of

\[ \frac{\tan^2\left(\frac{\pi}{7}\right) + \tan^2\left(\frac{2\pi}{7}\right) + \tan^2\left(\frac{3\pi}{7}\right)}{\cot^2\left(\frac{\pi}{7}\right) + \cot^2\left(\frac{2\pi}{7}\right) + \cot^2\left(\frac{3\pi}{7}\right)} = \frac{21}{5}. \]